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Scrat [10]
4 years ago
6

A middle school wants to have its students see a movie at a local theater.The total cost of the movie and theater rental is $ 1,

500. what equation shows the how the the cost per student depends on the number who attend
Mathematics
1 answer:
vlabodo [156]4 years ago
5 0

Answer: x = 1500/ n

Step-by-step explanation:

X- price per student

N- number of students

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vichka [17]

Answer: 135

Step-by-step explanation: 3x45=135

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Help pls I will give BRAINLY
OlgaM077 [116]

Answer: The missing length is 16/3

Step-by-step explanation:

First, you have to find the proportional value between the two lengths on the first figure and the two lengths on the second figure.

The first figure’s lengths are 8 and 9, so the shorter length is 8/9 of the longer length.

Now apply the same proportional value to the second figure.

6 * 8/9 = 48/9

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Which number is not equal to one of the following expressions? 2,500 × 10 250,000 ÷ 100 2,500 ÷ 10
ololo11 [35]

Answer:

2,500 ÷ 10

Step-by-step explanation:

Let's evaluate each expression one after the other and find out what they both gives us, then we'd now pick or select the one that doesn't give the required answer.

For 2,500 × 10

2500*10 = 25000

For 250,000 ÷ 10

250000/100= 25000

For 2,500 ÷ 10

2500/10= 250

So from the look of the results,

The one that gave us the odd result is

2,500 ÷ 10, and that's our answer.

4 0
3 years ago
Who can help me d e f thanks​
12345 [234]

d)

y = (2ax^2 + c)^2 (bx^2 - cx)^{-1}

Product rule:

y' = \bigg((2ax^2+c)^2\bigg)' (bx^2-cx)^{-1} + (2ax^2+c)^2 \bigg((bx^2-cx)^{-1}\bigg)'

Chain and power rules:

y' = 2(2ax^2+c)\bigg(2ax^2+c\bigg)' (bx^2-cx)^{-1} - (2ax^2+c)^2 (bx^2-cx)^{-2} \bigg(bx^2-cx\bigg)'

Power rule:

y' = 2(2ax^2+c)(4ax) (bx^2-cx)^{-1} - (2ax^2+c)^2 (bx^2-cx)^{-2} (2bx - c)

Now simplify.

y' = \dfrac{8ax (2ax^2+c)}{bx^2 - cx} - \dfrac{(2ax^2+c)^2 (2bx-c)}{(bx^2-cx)^2}

y' = \dfrac{8ax (2ax^2+c) (bx^2 - cx) - (2ax^2+c)^2 (2bx-c)}{(bx^2-cx)^2}

e)

y = \dfrac{3bx + ac}{\sqrt{ax}}

Quotient rule:

y' = \dfrac{\bigg(3bx+ac\bigg)' \sqrt{ax} - (3bx+ac) \bigg(\sqrt{ax}\bigg)'}{\left(\sqrt{ax}\right)^2}

y'= \dfrac{\bigg(3bx+ac\bigg)' \sqrt{ax} - (3bx+ac) \bigg(\sqrt{ax}\bigg)'}{ax}

Power rule:

y' = \dfrac{3b \sqrt{ax} - (3bx+ac) \left(-\frac12 \sqrt a \, x^{-1/2}\right)}{ax}

Now simplify.

y' = \dfrac{3b \sqrt a \, x^{1/2} + \frac{\sqrt a}2 (3bx+ac) x^{-1/2}}{ax}

y' = \dfrac{6bx + 3bx+ac}{2\sqrt a\, x^{3/2}}

y' = \dfrac{9bx+ac}{2\sqrt a\, x^{3/2}}

f)

y = \sin^2(ax+b)

Chain rule:

y' = 2 \sin(ax+b) \bigg(\sin(ax+b)\bigg)'

y' = 2 \sin(ax+b) \cos(ax+b) \bigg(ax+b\bigg)'

y' = 2a \sin(ax+b) \cos(ax+b)

We can further simplify this to

y' = a \sin(2(ax+b))

using the double angle identity for sine.

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33 degrees

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= 0.5795 rad = 33.203° = 33°12'11"

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